Question

# The smallest 5-digit number exactly divisible by 41 isa) 1004b) 10004c) 10045d) 10025

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Hint: In this question we need to find the greatest 4 digit number divisible by 41. Thus, by adding 41 to it will get the smallest 5 digit number divisible by 41.

$\text{Dividend=Quotient}\times \text{Divisor + Remainder }......\text{(i)}$
\begin{align} & \text{Dividend-Remainder=Quotient}\times \text{Divisor }......\text{(ii)} \\ & \Rightarrow \dfrac{\text{Dividend-Remainder}}{\text{Divisor}}=\text{Quotient} \\ \end{align}
The smallest n digit number divisible by an integer x =$({{10}^{n}}-\text{remainder obtained by dividing 1}{{\text{0}}^{n}}\text{by x)+x}$